Prove by induction that for all $n\in\mathbb N, (\sqrt3+i)^n+(\sqrt3-i)^n=2^{n+1}\cos(\frac{n\pi}6)$

Question

I want to prove by induction that for all $n \in \mathbb{N}$, $$(\sqrt{3} + i)^n + (\sqrt{3} - i)^n = 2^{n+1} \cos\left(\frac{n\pi}{6} \right)$$

I can prove the identity using direct complex number manipulation and de Movire (which gets it for all integers), but I'm getting stuck when trying induction out of interest instead.

For the inductive step, I have $$(\sqrt{3} + i)^{k+1} + (\sqrt{3} - i)^{k+1} $$ but what can I do with that? If I separate the terms into $(\sqrt{3} + i)^{k+1} = (\sqrt{3} + i)^{k} (\sqrt{3} + i)$ etc, the minus sign from the second term is causing me troubles.

Answer 1

You need to induct the two statements below simultaneously

$$(\sqrt{3} + i)^n + (\sqrt{3} - i)^n = 2^{n+1} \cos\frac{n\pi}{6} \\ (\sqrt{3} + i)^n - (\sqrt{3} - i)^n = i 2^{n+1} \sin\frac{n\pi}{6} $$

Note

\begin{align} &(\sqrt{3} + i)^{n+1}+ (\sqrt{3} - i)^{n+1} \\ =& \sqrt3[ (\sqrt{3} + i)^{n}+ (\sqrt{3} - i)^{n}]+i [(\sqrt{3} +i)^{n}- (\sqrt{3} - i)^{n}]\\ =& 2^{n+2} \left( \frac{\sqrt3}2\cdot\cos\frac{n\pi}{6} - \frac12\cdot\sin\frac{n\pi}{6} \right)\\ =& 2^{n+2} \cos\frac{(n+1)\pi}{6} \end{align}

Similarly

\begin{align} &(\sqrt{3} + i)^{n+1}-(\sqrt{3} - i)^{n+1}=i 2^{n+2} \sin\frac{(n+1)\pi}{6} \end{align}

Answer 2

Well, a possible way for you is to consider writing

$$z_n=(\sqrt 3 +i)^n + (\sqrt 3 -i)^n$$

Then can you go on to show that $z_{n}=2\sqrt3 z_{n-1}-4z_{n-2}$?

And then go on to consider $z_{n+1}$? Whence, by induction, $z_n$ can be shown to be real for all $n$.

Answer 3

My try is not that 'pure' in the sense of using the trig. equation.

Let $ A = \sqrt 3 + i$, $B = \sqrt 3 - i$. With some initial step verified, we can write \begin{align*} A^{n+1} + B^{n+1}&= (A^n + B^n)( A+ B) - AB (A^{n-1} + B^{n-1}) \\ & = 2^{n+2}\left( 2 \cos(\frac{n\pi}{6})\cos(\frac{\pi}{6}) - \cos \frac{(n-1)\pi}{6} \right) \\ & = 2^{n+2} \cos \frac{(n+1)\pi}{6} \end{align*} with $AB = (\sqrt3)^2 - i^2 =4$.